Fast convergences towards Euler-Mascheroni constant - SciELO

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Volume 29, N. 3, pp. 479–491, 2010 Copyright © 2010 SBMAC ISSN 0101-8205 www.scielo.br/cam

Fast convergences towards Euler-Mascheroni constant CRISTINEL MORTICI

Valahia University of Târgovi¸ste, Department of Mathematics, Bd. Unirii 18, 130082 Târgovi¸ste, Romania E-mails: [email protected] / [email protected]

Abstract. The aim of this paper is to introduce a new family of sequences which faster converge to the Euler-Mascheroni constant. Finally, numerical computations are given.

Mathematical subject classification: 41A60, 41A25, 57Q55. Key words: Euler-Mascheroni constant; speed of convergence. 1 Introduction

One of the most important constants in mathematics is defined as the limit of the sequence 1 1 1 γn = 1 + + + ∙ ∙ ∙ + − ln n, 2 3 n denoted γ = 0.57721566490153286 . . . . It is now known as the Euler-Mascheroni constant, in honour of the Swiss mathematician Leonhard Euler (17071783) and of the Italian mathematician Lorenzo Mascheroni (1750-1800). The sequence (γn )n≥1 and the constant γ have numerous applications in many areas of mathematics, such as analysis, theory of probability, special functions, and number theory. As a consequence, many authors are preoccupied to improve the speed of convergence of the sequence (γn )n≥1 , which is very slowly, if we take into account that it converges toward its limit like n −1 . #CAM-122/09. Received: 31/VIII/09. Accepted: 29/VII/10.

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FAST CONVERGENCES TOWARDS EULER-MASCHERONI CONSTANT

More precisely, we mention the following results related to the speed of convergence of the sequence (γn )n≥1 : 1 1 (Young) < γn − γ < 2 (n + 1) 2n

(see, [14, 15, 28]). We also refer here to the papers [1, 2, 5-12, 20-27], where important improvements of the speed of convergence of γn were established. The complete asymptotic expansion of the sequence (γn )n≥1 is X B2k 1 1 γn ∼ γ + − , 2n k=1 2k n 2k ∞

where the Bernoulli numbers B2k are defined by

X Bk t = tk. et − 1 k! k=0 ∞

As B2 = 1/6, B4 = −1/30, B6 = 1/42, B8 = −1/30, we imply γn ∼ γ +

1 1 1 1 1 − + − + − ∙∙∙ . 2n 12n 2 120n 4 252n 6 240n 8

DeTemple [3-4] introduced the sequence Rn = 1 +

  1 1 1 1 + + ∙ ∙ ∙ + − ln n + 2 3 n 2

which converges to γ like n −2 , since

1 1 < Rn − γ < . 2 24n 2 24 (n + 1)

Recently, Mortici [17] introduced the sequences

and

  1 1 1 1 1 un = 1 + + + ∙ ∙ ∙ + + √  − ln n + √ 2 3 n−1 6 6−2 6 n

  1 1 1 1 1 vn = 1 + + + ∙ ∙ ∙ + + √  − ln n − √ , 2 3 n−1 6 6+2 6 n

Comp. Appl. Math., Vol. 29, N. 3, 2010

(1)

(2)

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which converges as n −3 , since

6 lim n (u n − γ ) = − n→∞ 108 3

√

481

lim n (vn − γ ) =

and

n→∞

3

6 . 108 √

See [17, Theorem 2.1]. Furthermore, the arithmetic mean of the sequences (u n )n≥1 and (vn )n≥1 ,   1 1 1 1 1 1 2 zn = 1 + + + ∙ ∙ ∙ + + − ln n − 2 3 n − 1 2n 2 6

converges to γ as n −4 . We open here a new direction to accelerate the sequence (γn )n≥1 , that is to consider an additional term of the form Mn = γn − γ + ln

P (n) , Q (n)

where P, Q are polynomials of the same degree, having the leading coefficient equal to one. Precisely, we introduce the sequences νn = γn + ln

n−

n+

1 12 5 12

and

μn = γn + ln

33 37 n + 1680 140 103 61 n 2 + 140 n + 336

n2 +

whose speeds of convergence increase to n −3 , respective n −5 , since lim n 3 (νn − γ ) =

n→∞

−7 288

and

lim n 5 (μn − γ ) =

n→∞

3959 . 806 400

Our study is based on the following result, which represents a powerful tool for constructing some asymptotic expansions, or to accelerate some convergences. Lemma 1.

If (ωn )n≥1 is convergent to zero and there exists the limit lim n k (ωn − ωn+1 ) = l ∈ [−∞, ∞] ,

n→∞

with k > 1, then there exists the limit:

lim n k−1 ωn =

n→∞

l . k−1

(3)

For proofs and further applications, see [13-19]. The sequences (1)-(2) were introduced in [17] using Lemma 1. Clearly the sequence (ωn )n≥1 converges more quickly when the value of k satisfying (3) is larger. Comp. Appl. Math., Vol. 29, N. 3, 2010

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FAST CONVERGENCES TOWARDS EULER-MASCHERONI CONSTANT

2 First degree term

In this section we define the sequence

n+a n+b to find the values a, b which provide the fastest sequence (ωn )n≥1 . First ωn = γn − γ + ln

n+a n+1+a 1 n − ln + ln − ln , n+1 n+1 n+b n+1+b and we are concentrated to compute a limit of the form (3). In this sense, we used a computer software to obtain the following representation in power series:     1 1 2 1 2 2 ωn − ωn+1 = a − b + + −a − a + b + b − 2 n2 3 n3 (4)     3 3 3 1 1 + a 3 + a 2 + a − b3 − b2 − b + +O . 2 2 4 n4 n5 ωn − ωn+1 = −

We can state the following Theorem 2.

i) If a − b + 21 6= 0, then the speed of convergence of the sequence (ωn )n≥1 is n −1 , since lim n 2 (ωn − ωn+1 ) = a − b +

n→∞

1 2

and

lim nωn = a − b +

n→∞

1 6= 0. 2

ii) If a − b + 21 = 0 and −a 2 − a + b2 + b − 23 6= 0, then the speed of convergence of the sequence (ωn )n≥1 is n −2 , since and

lim n 3 (ωn − ωn+1 ) = −a 2 − a + b2 + b −

n→∞

2 3

  1 2 2 2 lim n ωn = −a − a + b + b − 6= 0. n→∞ 2 3 2

iii) If a − b + 21 = 0 and −a 2 − a + b2 + b − 23 = 0 (equivalent with a = −1/12, b = 5/12), then the speed of convergence of the sequence (ωn )n≥1 is n −3 , since lim n 4 (ωn − ωn+1 ) =

n→∞

Comp. Appl. Math., Vol. 29, N. 3, 2010

−7 96

and

lim n 3 ωn =

n→∞

−7 . 288

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483

The proof of Theorem 2 easily follows from Lemma 1 and (4). For a = −1/12, b = 5/12, the relation (4) becomes   7 1 ωn − ωn+1 = − +O 4 96n n5

and so, iii) is completely proved. 3 Second degree term

Now we define the sequence

n 2 + an + b n 2 + cn + d to find the values a, b, c, d which provide the fastest sequence (λn )n≥1 . First λn = γn − γ + ln

λn − λn+1 = −

n 2 + an + b 1 n − ln + ln 2 n+1 n+1 n + cn + d

(n + 1)2 + a (n + 1) + b , (n + 1)2 + c (n + 1) + d and we are concentrated to compute a limit of the form (3). In this sense, we used again the computer software to obtain the following representation in power series:     1 1 2 1 2 2 λn − λn+1 = a − c + − a + a − c − c − 2b + 2d + 2 n2 3 n3   3 3 3 1 + a − 3b − c + 3d − 3ab + 3cd + a 2 + a 3 − c2 − c3 + 2 2 4 n4  − a 4 + 2a 3 − 4a 2 b + 2a 2 − 6ab + a + 2b2 − 4b − c4  4 1 3 2 2 2 (1) −2c + 4c d − 2c + 6cd − c − 2d + 4d + 5 n5  5 10 5 5 + a − 5b − 5a 3 b + 5d − c − c2 − c3 − c4 − c5 + a 4 2 3 2 2 − ln

+10cd − 5d 2 + 10c2 d + 5c3 d − 5cd 2 + a 5 + 5ab2    10 3 5 2 5 1 1 2 2 −10ab + 5b + a − 10a b + a + +O 7 . 6 3 2 6 n n

We can state the following

Comp. Appl. Math., Vol. 29, N. 3, 2010

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FAST CONVERGENCES TOWARDS EULER-MASCHERONI CONSTANT

Theorem 3.

i) Let us denote the coefficients of (5) by

1 α =a−c+ 2   2 2 2 β = − a + a − c − c − 2b + 2d + 3 3 3 3 δ = a − 3b − c + 3d − 3ab + 3cd + a 2 + a 3 − c2 − c3 + 2 2 4  η = − a 4 + 2a 3 − 4a 2 b + 2a 2 − 6ab + a + 2b2 − 4b − c4  4 3 2 2 2 − 2c + 4c d − 2c + 6cd − c − 2d + 4d + 5

ii) If α 6 = 0, then the speed of convergence of the sequence (λn )n≥1 is n −1 , since lim n 2 (λn − λn+1 ) = α and

n→∞

lim nλn = α 6= 0.

n→∞

iii) If α = 0 and β 6= 0, then the speed of convergence of the sequence (λn )n≥1 is n −2 , since lim n 3 (λn − λn+1 ) = β

and

lim n 4 (λn − λn+1 ) = δ

and

n→∞

lim n 2 λn =

β 6= 0. 2

lim n 3 λn =

δ 6= 0. 3

n→∞

iv) If α = β = 0 and δ 6= 0, then the speed of convergence of the sequence (λn )n≥1 is n −3 , since n→∞

n→∞

v) If α = β = δ = 0 and η 6= 0, then the speed of convergence of the sequence (λn )n≥1 is n −4 , since η lim n 5 (λn − λn+1 ) = η and lim n 4 λn = 6= 0. n→∞ n→∞ 4

vi) If a = β = δ = η = 0 (equivalent with a = 33/140, b = 37/1680, c = 103/140, d = 61/336), then the speed of convergence of the sequence (λn )n≥1 is n −5 , since lim n 6 (λn − λn+1 ) =

n→∞

Comp. Appl. Math., Vol. 29, N. 3, 2010

3959 161 280

and

lim n 5 λn =

n→∞

3959 . 806 400

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The proof of Theorem 3 easily follows from Lemma 1 and (5). For a = 33/140, b = 37/1680, c = 103/140, d = 61/336, the relation (5) becomes   3959 1 . λn − λn+1 = +O 6 161 280n n7 and so, vi) is completely proved. 4 Concluding remarks

As least theoretically, further sequences of the form Mn = γn − γ + ln

P (n) Q (n)

can be defined, where deg P = deg Q = k ≥ 3. As above, Mn − Mn+1 = −

1 n P (n) Q (n + 1) − ln + ln Q (n) P (n + 1) n+1 n+1

(6)

and if we expand (6) into a power series of n −1 , then the 2k coefficients of the polynomials P and Q are the unique solution of the system obtained by imposing that the first 2k coefficients of the power series (6) vanish. In this case,   θ 1 Mn − Mn+1 = 2k+2 + O , n n 2k+3 with θ 6= 0. By Lemma 1, (Mn )n≥1 tends to zero as n −(2k+1) , since lim n 2k+1 Mn =

n→∞

θ . 2k + 1

Finally, we offer some numerical computations to prove the superiority of our sequences (νn )n≥1 and (μn )n≥1 over the classical sequence (γn )n≥1 and the DeTemple sequence (Rn )n≥1 . Remark that already μ1 approximates γ with seven exact decimals. n

γn − γ

Rn − γ

γ − νn

10

4. 916 7 × 10−2

3. 773 3 × 10−4

2. 274 8 × 10−5

100

4. 991 7 × 10−3

4. 125 2 × 10−6

2. 414 7 × 10−8

50

300

9. 966 7 × 10−3 1. 665 7 × 10−3

1. 633 7 × 10−5

4. 614 2 × 10−7

Comp. Appl. Math., Vol. 29, N. 3, 2010

1. 919 2 × 10−7

8. 982 5 × 10−10

μn − γ

4. 323 7 × 10−8

1. 533 0 × 10−11

4. 849 9 × 10−13

2. 012 2 × 10−15

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FAST CONVERGENCES TOWARDS EULER-MASCHERONI CONSTANT

Our approach works by small degrees k of P (t) and Q (t) , respectively, compared with the obtained order 2k + 1 of the speed of convergence, but for larger values of k, it becomes quite difficult (even for computer softwares) to compute the coefficients of the polynomials and the limit θ/ (2k + 1) of the sequences n 2k+1 Mn . We propose now a refined numerical method to compute the coefficients based on an explicit formula for the series expansion of Mn − Mn+1 around 1/n. Starting with formula (6), we have Mn − Mn+1 = −

1 n P (n) P (n + 1) − ln + ln − ln . n+1 n+1 Q (n) Q (n + 1)

(7)

As in the cases studied above, we assume that the polynomials P and Q have rational coefficients. Let us consider the factorizations P (t) = t k + ak−1 t k−1 + ∙ ∙ ∙ + a0 = (t − α1 ) (t − α2 ) ∙ ∙ ∙ (t − αk )

Q (t) = t k + bk−1 t k−1 + ∙ ∙ ∙ + b0 = (t − β1 ) (t − β2 ) ∙ ∙ ∙ (t − βk ) ,

(8) (9)

where the roots αν and βν are complex numbers. Then (7) takes the form
! k X n − αj − 1 n − αj 1 n
. (10) Mn −Mn+1 = − −ln + ln − ln n − βj n+1 n + 1 j=1 n − βj − 1 Next, we have for real numbers α, β that   X   ∞ n−α β −α (−1)ν−1 β − α ν ln = ln 1 + = n−β n−β ν n−β ν=1   ∞ X 1 (−1)ν−1 β − α ν = ν n (1 − β/n)ν ν=1 !  ∞ ∞ X β μ (−1)ν−1 (β − α)ν X ν + μ − 1 = νn ν n ν−1 ν=1 μ=0 ! ! ∞ d X X d − 1 (α − β)ν β d−ν 1 =− (with d = ν + μ) ν nd ν−1 d=1 ν=1 ∞ X αd − β d =− . dn d d=1

Comp. Appl. Math., Vol. 29, N. 3, 2010

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487

Replacing α by α − 1 and β by β − 1, we get

∞ X n − (α − 1) (α − 1)d − (β − 1)d =− ln n − (β − 1) dn d d=1

and

(11)

X (−1)d n ln = . n+1 dn d d=1

(12)

∞

Finally, using

X (−1)d−1 1 = , nd n+1 d=1

(13)

∞

it follows from (10)-(13):

Mn − Mn+1 = 

× (−1)d (d − 1) +

k  X j=1

αj − 1


d

∞ X d=2

− α dj



βj − 1


d

− β dj





 1 . dn d

(14)

For every d the corresponding term in brackets on the right-hand side of (14) is a polynomial in 2k variables, which is symmetric in α1 , α2 , . . . , αk and in β1 , β2 , . . . , βk . Therefore, the proof of the main theorem on symmetric polynomials involves an algorithm to express the polynomials (14) in terms of the elementary symmetric polynomials in α1 , α2 , . . . , αk and in β1 , β2 , . . . , βk . So, one obtains equations in terms of the coefficients a0 , . . . , ak and b0 , . . . , bk . But, a second method to deduce the rational coefficients a0 , . . . , ak and b0 , . . . , bk from the (complex) solution of the system

×



αj − 1


d

(−1)d (d − 1) + − α dj



βj − 1


d

k X j=1

− β dj = 0 (2 ≤ d ≤ 2k + 1) 

(15)

is based on the numerical continued fraction algorithm. Provided that the coefficients a0 , . . . , ak−1 and b0 , . . . , bk−1 are rationals, the following pure numerical method works without using any computer algebra software in order to obtain the coefficients. Here, we first solve the system (15) numerically, Comp. Appl. Math., Vol. 29, N. 3, 2010

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FAST CONVERGENCES TOWARDS EULER-MASCHERONI CONSTANT

which for large k is much simpler than to compute the coefficients a0 , . . . , ak and b0 , . . . , bk by a computer algebra system. We demonstrate the method by computing the rationals a0 , a1 , a2 and b0 , b1 , b2 for k = 3. The system 0=1+

3  X j=1

3  X

0 = −2 + 0=3+

j=1

3  X j=1

j=1

3  X j=1

0 = −6 +

− α 2j

αj − 1
4

j=1


6

αj − 1



βj − 1



βj − 1



βj − 1


3

− α 3j


5

− α 5j


7

− α 7j

− α 4j

αj − 1

αj − 1

3  X

has (the unique) solution


2

αj − 1

3  X

0 = −4 + 0=5+

αj − 1

− α 6j


2

− β 2j



βj − 1



βj − 1



βj − 1


3

− β 3j


5

− β 5j


7

− β 7j


4


6



− β 4j

− β 6j











α1 = −0.25815871587916770707043092744853466448964330026005...

+i ∙ 0.48397242106377239253881674751055982149144184289297...

α2 = 0.038201465324611971071624393136243082171731381749928...

α3 = −0.25815871587916770707043092744853466448964330026005...

−i ∙ 0.48397242106377239253881674751055982149144184289297...

β1 = −0.38406133637441025553569425604004004944633805438883... β2 = −0.29702731502965659376677160286039309868060858219067...

−i∙ ∙ 0.49547669307316290426997439537218740680230881850428...

β3 = −0.29702731502965659376677160286039309868060858219067...

+i ∙ 0.49547669307316290426997439537218740680230881850428...

Comp. Appl. Math., Vol. 29, N. 3, 2010

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489

It follows that −a2 = α1 + α2 + α3

= −0.47811596643372344306923746176082624680755521877097...

a 1 = α1 α 2 + α 2 α 3 + α 3 α 1

= 0.28115114446890147824727263979600428198559039680544...

−a0 = α1 α2 α3

= 0.011493874548781090837165603520743707659595510062767...

−b2 = β1 + β2 + β3

= −0.97811596643372344306923746176082624680755521877081...

b1 = β1 β2 + β2 β3 + β3 β1

= 0.56187579435242986644855803734308407205603467285750...

−b0 = β1 β2 β3

= −0.12816986295374145841435561061729286028351448912202...

Next, the continued fraction algorithm applied to the above numerical values, gives 68143 142524 186997 = < 0, 3, 1, 1, 3, 1, 9, 10, 3, 1, 3, 1, 10 > = 665112 74381 = − < 0, 87, 349, 4, 1, 5, 3, 2 > = − 6471360 139405 = − < −1, 45, 1, 2, 3, 1, 1, 7, 2, 2, 3 > = 142524 1121131 = < 0, 1, 1, 3, 1, 1, 5, 1, 2, 3, 1, 1, 8, 1, 21, 1, 3 > = 1995336

a2 = − < −1, 1, 1, 10, 1, 12, 7, 1, 1, 5, 1, 4 > = a1

a0 b2 b1

b0 = − < −1, 1, 6, 1, 4, 18, 2, 3, 1, 1, 4, 1, 8, 1, 5, 3, 1, 1, 2, 3 > =

Then we have

Comp. Appl. Math., Vol. 29, N. 3, 2010

lim n 7 Mn =

n→∞

θ 7

30689033 . 239440320

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FAST CONVERGENCES TOWARDS EULER-MASCHERONI CONSTANT

with θ=



1 7+ 8

3  X j=1

αj − 1


8

− α 8j



βj − 1


8





− β 8j 

= −0.01832772866046807884381439244713734156447898226906...

=< −1, 1, 53, 1, 1, 3, 1, 1, 10, 7, 1, 1, 1, 3, 2, 3, 4, 1, 2, 1, 36, 1, 8, 1, 3, 3 > 10833071983 =− . 591075532800

It is to be noticed that these results were rediscovered by us using Lemma 1 presented in the first part of this paper. We omit the proof for sake of simplicity. Finally, remark that if the polynomials P and Q of k-th degree are already determined, say using the previous numerical method, then the problem of verifiying the speed of convergence of the corresponding sequence Mn using Lemma 1 becomes a much easier task. Acknowledgements. The author thanks the editors and the anonymous referee for useful ideas which improved much the initial form of this paper. REFERENCES

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